Why a Pulley Creates a Rate Constraint
A rope in a pulley system has a fixed total length, provided it does not stretch and remains taut. When a load moves, some portions of the rope get shorter or longer. The free end must move in a way that keeps the total length unchanged.
For a common movable-pulley arrangement, two rope segments support the moving pulley attached to the load. If the load rises by a certain distance, both supporting segments shorten by that distance. The pulled end must therefore supply twice that distance of rope. This factor of two is the central feature of the constraint.
We will use a particular arrangement and define its coordinates carefully. The signs of the rates will then tell us whether the load or the free end is moving up or down.
The Arrangement and Its Coordinates
Consider a rope with one end fixed to an overhead support. It runs down around a movable pulley attached to the load, then back up over a fixed pulley and down to the free end. Assume the two rope segments supporting the movable pulley remain vertical and that the pulleys and rope are ideal: the rope does not stretch, and the rope length around each pulley is constant.
Let \(y(t)\) be the height of the load, measured upward from a fixed reference level. Let \(x(t)\) be the length of the hanging rope from the fixed pulley to the free end, measured downward. Thus, an increase in \(y\) means the load rises, while an increase in \(x\) means the free end moves downward and more rope hangs below the fixed pulley.
The overhead supports remain fixed, so each of the two supporting segments has length \(H-y(t)\), where \(H\) is a fixed constant determined by the chosen reference level and geometry. All fixed rope portions and the lengths wrapped around the pulleys can be combined into another constant, \(C\).
This equation applies to the specified arrangement. A different routing of the rope can have a different number of changing segments, and therefore a different constraint.
Deriving the Speed Relation
Since the rope length \(L\) is constant, differentiate the constraint with respect to time. The constants \(H\), \(C\), and \(L\) have derivative zero. By the chain rule, the derivative of \(y(t)\) is \(dy/dt\), and the derivative of \(x(t)\) is \(dx/dt\).
Rearranging gives the rate relation:
The factor of two comes from the two supporting segments: a change in load height changes both of them. The sign follows from our coordinate choices. If the load rises, \(dy/dt>0\), then \(dx/dt>0\), so the free end moves downward. If the free end is pulled upward, then \(dx/dt<0\), and the load moves downward, so \(dy/dt<0\).
The signed equation is more informative than a statement about speeds alone. Speeds are nonnegative magnitudes, so they do not record direction. The equation with signed rates does.
A Reliable Method
In a pulley problem, the main task is to count and describe the changing rope segments before differentiating. The following sequence keeps the geometry and signs connected:
State which direction counts as positive for the load and for the free end.
Write the length of each changing portion in terms of the chosen coordinates. Put fixed lengths into constants.
Set the sum of all rope portions equal to the constant total length.
Apply the chain rule to each changing coordinate and retain the signs.
Isolate the requested rate, state its direction, check the units, and substitute it into the differentiated constraint.
Worked Examples
Worked Example: Finding the Load’s Speed
The load in this arrangement rises at \(0.75\) metres per second. Find the free end’s signed rate and describe its motion.
Use the established constraint and differentiate it:
The load is rising, so \(dy/dt=0.75\ \text{m/s}\). Substitute this value:
Because \(x\) is measured downward, the positive rate means the free end moves downward at \(1.50\ \text{m/s}\). Check by substitution: \(-2(0.75)+1.50=0\), so the differentiated length constraint is satisfied. The units also agree: both terms are rates of length, in metres per second.
Worked Example: Finding the Load’s Speed from the Pulled-End Speed
The free end is pulled downward at \(2.4\ \text{m/s}\). Find the load’s signed vertical rate.
Downward is positive for \(x\), so \(dx/dt=2.4\ \text{m/s}\). From the differentiated constraint,
Substitute the known rate and solve:
The result is positive, so the load rises at \(1.2\ \text{m/s}\). Check in the original rate equation: \(2(1.2\ \text{m/s})=2.4\ \text{m/s}=dx/dt\). The pulled end moves twice as fast as the load.
Worked Example: The Free End Is Pulled Upward
The free end is pulled upward at \(0.80\ \text{m/s}\). Find the load’s motion and check the result using a short displacement.
Since \(x\) is measured downward, an upward motion makes \(x\) decrease. Therefore \(dx/dt=-0.80\ \text{m/s}\). Solve the rate constraint for the load:
The negative sign means the load descends at \(0.40\ \text{m/s}\). To check the geometry, suppose this motion continues for \(0.25\) seconds. The free end moves upward by \(0.80(0.25)=0.20\) metres, so its segment changes by \(\Delta x=-0.20\ \text{m}\). The load descends by \(0.40(0.25)=0.10\) metres, so \(y\) changes by \(-0.10\ \text{m}\). Each of the two supporting segments then lengthens by \(0.10\) metres, for a combined increase of \(0.20\) metres. The total change in rope length is
The total rope length remains unchanged, as required. The signed-rate check gives \(-2(-0.40)-0.80=0\), also satisfying the differentiated constraint.
Common Mistake: Counting Only One Supporting Segment
A frequent error is to treat the moving load as if only one rope segment changes length. That leads to the incorrect equation \(L=(H-y)+x+C\), and differentiating it gives \(dx/dt=dy/dt\). For example, this wrong equation would say that a load rising at \(0.75\ \text{m/s}\) requires the free end to move down at only \(0.75\ \text{m/s}\).
But the movable pulley is supported by two changing segments. When the load rises \(0.75\) metres, each segment shortens by \(0.75\) metres, and together they release \(1.50\) metres of rope. The free-end segment must lengthen by \(1.50\) metres to keep the total fixed. The corrected equation is \(L=2(H-y)+x+C\), which gives \(dx/dt=2\,dy/dt\). The error is geometric: the number of supporting rope segments was counted incorrectly.
A related sign mistake is to call both directions positive without defining what the coordinates measure. With the conventions here, upward motion of the load makes \(y\) increase, while upward motion of the free end makes \(x\) decrease. Keeping those definitions beside the equation prevents the direction from being lost.
Why the Constraint Matters
The rope-length equation does more than provide a numerical factor. It explains how a pulley trades distance for force in an ideal mechanical system: the load moves through half the distance that the pulled end moves. This tutorial establishes the kinematic relation between the motions; it does not require a force model to obtain that relation.
The relation depends on the geometry. In this arrangement, two segments change with the load’s position. If a system has a different number of supporting segments, its rope-length equation must reflect that geometry rather than reusing the factor of two automatically. Always build the constraint from the particular rope path.
For the arrangement studied here, the key result is the signed relation \(dx/dt=2\,dy/dt\). With the stated coordinate choices, a rising load corresponds to a downward-moving free end, and the free end’s speed is twice the load’s speed.
Check Your Understanding
Use the coordinate definitions and pulley arrangement from this tutorial.
- Why does the load’s position appear in two changing terms in the rope-length equation?
- If the load rises at \(0.60\ \text{m/s}\), what is \(dx/dt\), and which way does the free end move?
- If the free end is pulled upward at \(1.2\ \text{m/s}\), find the signed rate \(dy/dt\) and state the load’s direction.
- What incorrect speed relation results from counting only one supporting segment?
- Why must the sign of \(dx/dt\) be negative when the free end moves upward?
